THERMAL PROPERTIES OF MATTER

1) A sheet of copper has an area of 500cm² at 0˚C. find the area of this sheet at 80˚C.
The coefficient of area expansion is to good approximation 2α.thus

Answe:

∆A=2αA∆t
=2(1.67x10⁻⁵)(500)(80˚-0˚)
=1.34cm²

The new area is
500 cm²+1.34 cm²
=501.3cm²


2) If an anisotropic solid has coefficients of liner expansion αx, αy. Αz for three mutually perpendicular direction in the solid, what is the coefficient of volume expansion for the solid?

Answer:

Consider a cube,with edges parallel to X,Y,Z, of dimension L∘ at T=0. After a change in temperature ∆T=(T-0), the dimensions change to
Lx=L∘(1+αxT) Ly=L∘(1+αyT) Lz=L∘(1+αzT)
And the volume of the parallelepiped is
V=V∘(1+αxT)(1+αyT)(1+αzT)≈ V∘[1+(αx+αy+αz)T]
Where V∘=L∘³. Therefore, the coefficient of volume expansion is given by β≈αx+αy+αz. (note : the symbols β and γ are often used for the coefficient of value expansion.)


3) Calculate the increase in volume of 100cm³ of mercury when its temperature change from 10 to 35˚C.

Answer:

∆V=βV∘∆T. Then volume coefficient of expansion of mercury β=1.8x10⁻⁴˚C⁻¹. then ∆V=(1.8x10⁻⁴˚C⁻¹)(100cm³)(25˚C)=0.45cm³


4) The coefficient of liner expansion of glass. If a bottle holds 50.000cm³ at 15˚C, what is its capacity at 25˚C?

Answer:

The glass expands uniformly in all dimensions, so the capacity expands in proportion to the glass. The volume coefficient of expansion is related to the linear coefficient byβ=3α. Then V=V∘(1+3α∆T)=(50.000cm³)[1+3(8.3x10⁻⁶˚C⁻1)(10˚c)]. V=50.012cm³


5) Determine the change in volume of a block of iron 5cm x 10cm x 6cm, when the temperature change from 15˚C to 47˚C.

Answer:

V∘=300cm³; β=3α.then
∆V=3αV∘∆T=3(1.2 x 10⁻⁵˚C⁻¹)(300cm³)(32˚C)=0.35cm³



6) A glass vessel is filled with exactly 1L of turpentine at 20˚C. What volume of the liquid will overflow if the temperature is raised to 86˚C?

Answer:

As the temperature raises, both the capacity of the glass vessel and the volume of the turpentine increase. The amount of the overflow is the difference in these increases : ∆VT-∆VG=(βT-βG)V∆T. noting that βG=3αG, and using the data , we have
∆VT-∆VG=(9.4x10⁻⁴˚C⁻¹)-(0.25 x 10⁻⁴˚C⁻¹)(1L)(66˚C)=60mL


7) The density of gold is 19.30g/cm³ at 20˚C. compute the density of gold at 90˚C.

Answer:

d=M/V d1/d2=V2/V1
since mass constant. The indices 1 and 2 refer to 90˚C and 20˚C, respectively
d1/d2=V2/[V2(1+3α∆T)]
=1/[(1+3α∆T)]≈ 1-3α∆T (since 3α∆T<<1)
d1=d2(1-3α∆T)=(19.3g/cm³)[1-3(14.2 x 10⁻⁶˚C⁻¹)(70˚C)]
d1=19.24g/cm³


8) The density of mercury at 0˚C is 13600kg/m³.calculate the density of mercury at 50˚C.

Answer:

Let ρ1=ρ∘[V∘/V1]=ρ∘[V∘/(V∘+∆V)]=ρ∘[1/(1+(∆V/V∘)]
But ∆V/V∘=β∆T=(1.82 x 10⁻⁴˚C⁻¹)(50˚C)=0.0091
Substitution gives ρ1=(13600kg/m³)[1/(1+0.0091)]=13477kg/m³


9) the mercury in a thermometer has a volume of 210mm³ at 0˚C, at which temperature the diameter of the bore is 0.2mm. How far apart are the degree marks on the stem?

Answer:

We assume that the temperature rises from 0˚C to t. so ∆t=t. Then
∆VHg =β Hg V Hg ∆t, ∆VG = 3αGVG ∆t, where here VHg is the volume occupied by mercury at 0˚C and VG is the volume of the glass container occupied by the mercury at 0˚C. Indeed hAt = ∆VHg-∆VG, where h is the height that the mercury rises and At is the cross section of the bore at temperature t. At = ∏r(1+2αG∆t) with r=0.1mm, the radius at 0˚C, then
Sh=[(βHg-3αG)VHgt]/[ ∏r²(1+2αG t)]
=[ [(182-25) x 10⁻⁶˚C⁻1](210mm³)t]/[0.0314[1+(16.6 x 10⁻⁶˚C⁻¹)t]]


10) A glass beaker hold exactly 1L at 0˚C. (a) what is its volume at 50˚C? (b) If the beaker is filled with mercury at 0˚C, what volume of mercury overflow when the temperature is raised to 50˚C?

Answer:

a) The volume of the beaker after the temperature change is.
Vbeaker =V∘(1+3αG∆T)=(1)[1+3(8.3 x 10⁻⁶)(50)]=1.001L.
b)for the mercury expansion,
Vmercury=V∘(1+β∆T)=(1)[1+(1.82 x 10⁻⁴)(50)]=1.009L
The overflow is thus 1.009-1.001=0.008L, or 8mL

THERMAL PROPERTIES OF MATTER

1. A pie plate is filled up to the brim with pumpkin pie filling. The pie plate is made of Pyrex and its expansion can be neglected. It is a cylinder with an inside depth of 2.10 cm and an inside diameter of 30.0cm. It is prepared at a room temperature of 68 degrees Fahrenheit and placed in an oven at 400 degrees Fahrenheit. When it’s taken out, 151cc of the pie filling has flowed out and over the rim. Determine the coefficient of volume expansion of the pie filling, assuming it is a fluid.

Answer:

The volume expanded is 151 cc. the volume of reference, that is, the V of cylinder is 0.25*3.1415*2.10*(30.0) ^2the coefficient of volume expansion= 151/ VThe expansion of the Pyrex plate (container) is neglected.So V (final) =V (Initial)=V of the cylinder.

2. At the bottom of an old mercury-in-glass thermometer is a 46.0-mm3 resevoir filled with mercury. When the thermometer was placed under your tongue, the warmed mercury would expand into a very narrow cylindrical channel, called a capillary, whose radius was 2.50 x 10-2 mm. Marks were placed along the capillary that indicated the temperature. Ignore the thermal expansion of the glass and determine how far (in mm) the mercury would expand into the capillary when the temperature changed by 1.0 C°

Answer:

The change in the mercury's volume is equal to beta*V*delta T, where beta is the coefficient given above, V is the original volume, and delta T is the temperature change. Since the diameter of the capillary is unchanged, the change in volume there will be pi*r-squared*delta L, where r is the radius, and deltaL is the change in the length of the mercury column. Setting the volume changes equal to each other, and doing a little algebra,beta*V*deltaT = pi*r-squared*deltaLdeltaL = beta*V*deltaT / pi*r-squareddeltaL = 1.8x10(-4)*46*1.0 / 3.14*[2.5x10(-2)]*[2.5x10(-2)]deltaL = 4.21 mm

3. The coefficient of volume expansion for ethyl alcohol is 1.12x10^-3 1/K. Copper has a coefficient of volume expansion of 1.67x10^-3 1/K. If a copper cylinder is filled to the brim with ethyl alcohol at 5 C, what percentage of the alcohol will spill if the temperature is raised by 25 C.

Answer:

The temperature change is 20 C. So, the volume expansion of the copper will be20x1.67x10^-3 or 3.34x10^-2 liters and the volume expansion of the alcohol will be 20x1.12x10^-3=2.24x10^-2 l

4. The density of gasoline is 730 kg/m3 at 0°C. Its average coefficient of volume expansion is 9.60 x 10-4/C°. If 1.00 gallon of gasoline occupies 0.00380 m3, how many extra kilograms of gasoline would you get if you bought 8.0 gal of gasoline at 0°C rather than at 19.0°C from a pump that is not temperature compensated?

Answer:

Density of gasoline at 0°C = 730 kg/m3Density of gasoline at 19°C = 730 (1 + 19(9.60 x 10-4)) kg/m3 = 743.3152 kg/mm=dVV = 8 gal = 0.0304 m3difference in m = 743.3152(0.0304) - 730(0.0304) kg= 0.40478208 kg

5. A pair of eyeglass frames are made of an epoxy plastic (coefficient of linear expansion = 1.30 10-4°C-1). At room temperature (20.0°C), the frames have circular lens holes 2.10 cm in radius. To what temperature must the frames be heated if lenses 2.11 cm in radius are to be inserted into them?

ANSWER:

ΔL = αLoΔT(.0211 - .0210) = (1.3e-4)(.021)(T - 20).0001 = (2.73e-6)(T - 20)36.63003663 = T - 20T = 56.63003663T = 56.6°C

The frames must be heated to 56.6°C for the frames to increase from a 2.1cm radius to a 2.11cm radius.

6. The aptly-named Steel Bridge over the Willamette River in Portland, Oregon, has a central span steel truss that is 64.1 m long. What is its change in length over the year from the average minimum annual temperature of 0.90°C to the average maximum of 26.3°C? Use a coefficient of linear expansion of 1.17×10^−5 (1/C°).

Answer:

delta L = length chg of central truss = (L)(Cx) (delta T) = (64.1)(1.17e-5)(26.3 - 0.9)delta L = 75e-5(25.4) = 1905e-5 = 0.0191 m = 1.91 cm

7. We cut a circular piece from the rectangular plate. Which ones of the processes given below can help us in passing through the circular piece from the hole?

Answer:

I. Increasing the temperatures of rectangular plate and circular piece
II. Decreasing the temperature of the circular piece
III. Decreasing the temperatures of the rectangular plate and circular piece

I. If we increase the temperatures of the plate and circular piece, expansion of the hole and the circular piece will be the same. Thus, this option can help us.
II. If we decrease the temperature of the circular piece, it contracts and hole becomes larger than the piece. This option can also help us.
III. If we decrease the temperatures of the plate and circular piece, hole and circular piece contract in same size. This process can also help us.

8. There are three same metal rods having same length and thickness. If the temperatures of them are given like; T, 2T and 3T find the relations of final lengths of the rods.

Answer:

We find the final temperatures of the system by the formula;
Tfinal=T1+T2+T3/3=6T/3=2T
Since the temperature of the first rod increase, its final length also increases. Temperature of the second rod stays same, thus there won’t be change in the length of this rod. Finally, temperature of the third rod decreases, thus its contract and final length of it decreases with respect to initial length. As a result relation of the final lengths of the rods;
L1>L2>L3

week 7:electomagnetic wave

(week 6) wave and sound

Wave and sound

(1-0) types of wave

Water wave

Sound waves travel in air with a speed of 343m/s. The lowest frequancy sound we can hear is 20.0 Hz; the highest frequency is 20.0kHz. Find the wave length of sound for frequencies of 20.0Hz and 20.0kHz

Solution

Solve Equation for λ

λ = v/f = (343 m/s)/(20.0s⁻¹) = 17.2 m

λ = v/f = (343 m/s)/(20,000s⁻¹) = 1.72 m

(2.0) wave and string

The speed of a wave on a string

A 5.0-m length of rope, with a mass of 520g, is pulled taut with a tension of 46N. Find the speed of waves on the rope.

Solution

First, calculate the mass per length, μ:

520g = 0.52kg

μ = m/ L = (0.52kg)/(5.0m) = 0.10kg/m

Now, substitute μ and F into Equation (2.0):

ν = √(F/ μ) = √[46N/(0.10kg/m)]

(3.0) harmonic wave functions

A loudspeaker puts out 0.15 W of sound through a square area 2.0 m on each side. What is the intensity of this sound?

Solution

Applying Equation 14-5, with A = (2.0m)², we find

I = intensity = ?

P = power = 0.15W

A = area = (2.0m)²

I = P/A = (0.15W)/(2.0m)² = 0.038W/m²

(4.0) harmonic wave functions

The power of song

Two people relaxing on a deck listen to a songbird sing. One person, only 1.00m from the bird, hears the sound with an intensity of 2.80x10⁻⁶W/m². (a) what intensity is heard by the second person , who is 4.25m from the bird? Assume that no reflected sound is heard by either person. (b) What is the power output of the bird’s song?

Our problem, the two observers, one at a distance of r1 = 1.00m from the bird, the other at a distance of r2 = 4.25m. The sound emitted by the bird is assumed to spread out spherically, with no reflections

Strategy

a. the two intensities are related by equation 4.0 ,with r1 = 1.00m and r2 = 4.25m.

b. the power output can be obtained from the definition of intensity, I =( P/A). we can calculate P for either observer, noting that A = 4πr²

solution

part (a)

1. Substitute numerical values into equation (4.0)

I2 = (r1/r2)²I1 = (1.00m/4.25m)²(2.80 X 10⁻⁶ W/m²)

= 1.55 X 10⁻⁷ W/m²

Part (b)

2. Solve I =( P/A) for the power ,p, using data for observer 1:

I1 =( P/A1)

P = I1 A1 = (2.80 X 10⁻⁶W/m²)[4π(1.00m)²]

= 3.52 X 10⁻⁵W

3. As a check, report the calculation for observer 2:

I2 =( P/A2)

P = I2A2 = (1.55 X 10⁻⁷W/m²)[4π(4.25m)²]

= 3.52 X 10⁻⁵W

(5.0)

At a distance of 5m from a source the sound level is 90dB. How far away has the level dropped to 50dB

Solution

I1 = (P/4πr1²) and I2 = (P/4πr2²), so (I2 /I1) = (r1²/r2²)

Β1 = 10 log (I1 /I0 ) = 90dB , so (I1 /I0 ) = 10⁹

similarly (I2 /I0 ) = 10⁵

thus (I2 /I1) = (10⁵/10⁹) = 10⁻⁴ = (r1²/r2²), so r2 = 10²r1 = 500m